# Full text of "Treatise On Analysis Vol-Ii"

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11    THE SPACES L1 AND L2        163

We remark that, given /e JS?i(X, /z), it is not always possible to find a
function/' equivalent to /and a sequence of functions/, e «?TR(X) such that
the sequence (/„(*)) converges to/'O) for allxeX (Problem 3).

We have already defined (13.10) the vector space -S?£(X, /*) of complex-
valued integrable functions. Likewise, we denote by <£ £(X, ju) (or <£ cOu), or
JS?£) the set of c0ra/?/ex-valued /i-measurable functions such that \f\2 is
ji-integrable, and we define N^/) and N2(/) by the formula (13.11.1) for
any complex-valued function/ Since a complex-valued function is measurable
if and only if ^?/and .//are measurable, and since

it follows that

#S(X,Ai) = JS? J(X,M) 0 fcS?2

All the properties of J\$?£(X, ju) proved above extend immediately to
JSf£(X,/i) Q? = l,2), and the Banach space JS? J(X, //) (also denoted by
^ c(j") or «^c) ^ defined exactly as in the real case.

Furthermore, we have

(13.11.7) Iff, g belong to -SfJ(X, /*) (resp. JS?£(X, ^)) then their product f g
belongs to 4fi(X, ji) (resp. Jif i(X, ju)). T/ze space Lj(X, ^) (resp. Lj(X, ji))
w « separable Hilbert space with respect to the Hermitian form

(13.11.7.1)              (/, g) h+(/|0) = tfWtfddrtx),

and the corresponding norm is N2(/).

The product/^ is measurable (13.9.8.1), hence the first assertion follows
from (13.9.13) and (13.11.2.2). The second follows from the first together
with the Fischer-Riesz theorem (13.11.4) and (13.11.6).

PROBLEMS

1. Let A be Lebesgue measure on I = [0,1 [. For each integer « = 2"+ &(()<;/:< 2"),
let/n be the function which is equal to 1 on the interval [k • 2~h, (k+ 1) -2~*[,
and zero elsewhere. Show that the sequence (/„) converges to 0 in mean and in square
mean, but that the sequence (/„(*)) does not converge for any x e I,nough to observe that for any # e ^ R(X; K) we have